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Does anyone know the density of air at 2,500 PSI? One other Question.

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CJ8879

Mechanical
Nov 17, 2004
2
Just curious if anyone knows what the density of air is at 2,500 PSI? I'm assuming 70 degrees F for my calcs at a standard elevation.

One other question. I have been asked to determine how long a high pressure bottle of compressed air with a rated volume of 300 ft^3 can supply a piece of equipment downstream of a regulator before running out of air at the required pressure.

I figure the easiest way to solve it is to just do a mass balance on it and the density into the regulator times a flow rate = density out of the regulator times a flow rate. Sound right?
 
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Seems like my thermodynamics book had a compressability table for air, which would give the answer you need- only I don't have that book here at the office.

If you remove the air slowly, so that the tank stays at constant temperature, then your approach should work. If you're removing air fast enough, temperature in the tank will drop, giving you lower pressures than what you would expect.
 
You can calculate the density of compressed air by a modified version of the ideal gas law;

PV x (molecular weight of the gas)/RT = d
 
Thanks for your help. I didn't think the perfect gas law would work at that pressure, but I suppose it will be fine for an estimate. I'll use it.
 
I did locate another thermo book. It shows that for nitrogen at 2500 PSI and 300K temperature, the compressibility factor Z is about 1.022. Z is defined as Pv/RT, where v is specific volume. In other words, the density is about 2% less than you would predict with ideal gas laws. I assume air would be similar to nitrogen in this respect.

I suppose you would need to dry the air as or after it's compressed, or you would have liquid water in there, too, which would throw the numbers off some.
 
From Perry VI, specific volumes, m3/kg:


280K 300K

150 bar 0.00529 0.00578
200 bar 0.00407 0.00446

Just interpolate for the desired T,P conditions. [smile]
 
If you assume: negligible flow and heat losses between the source pressure vessel and the pressure regulator,and Z=constant and "quasi-steady flow"
---then for an isothermal source, the rate of pressure decrease is directly proportional to the mass flow rate.
-----for an adiabatic tank, pv^k =constant which yields p/m^k= constant where p is source pressure and m the mass in the tank. k, the spec ht ration Knowing the flow rate from the regulator allows calculation of mass in the source as a function of time and therefore, the source pressure as a function of time.
 
Someone correct me if I'm wrong, but if you know the flow rate and downstream pressure at point of use, and the interior volume and pressure of the tank, you should be able to use Boyle's Law. Providing the pressure downstream of the reg is constant.
 
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